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# PP08 - Texas State University—San Marcos Ingram School of...

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Unformatted text preview: Texas State University—San Marcos Ingram School of Engineering EE3340: Fields and Waves \$ 0 L. u T i o u 5 Practice Problem 8: Gauss’s Law, Divergence, Potential NAME: DATE: For this problem, you will use Gauss’s law and the concept of divergence to solve the same problem two different ways. Then you will ﬁnd the potential everywhere. Assume there is a slab of charged material whose relative dielectric constant e, = 2.8, and whose uniform volume charge density pv = +1 mC M3. The slab extends to inﬁnity in the y- and z—directions, has a thickness T: 1 cm, and is centered on the x = 0 plane: \>\\ i X yl-‘l \\\o :44 1. Using Gauss’s Law and symmetry arguments, ﬁnd the electric ﬁeld E - ii x E x (x) everywhere. (Since the slab is uniform and inﬁnite in y and 2, there is no Ey or E, ﬁeld.) For your volume, imagine a cylindrical “pillbox” with an axis in the x—direction and put its ends at +x and -x. Again by symmetry, the ﬂux coming out each end has to be equal in magnitude but opposite in direction (sign). You will need three expressions for Ex: one for x < -T/2, one for -T/2 < x < +T/2, and one for x > T/2. When you are done, sketch your Ex(x) on the graph provided. Q5 -3 L n“ ( (0’1“) 1(1.%)(4.i5¥..°~u”,,} : 2.0 l lav/M aux . By 6x symmetry, there is no E)c at the center of the slab (x = 0). Find Dx and then E, by integrating the divergence equation from x = 0 to +x and -x values. Your result should be thesamethatyougotinPartl. x>°1 6 E 35* _. 54E _\___3_C_5_;3__- . ' ° T -‘ I“ "J x ‘0. sureSg‘“ +OK: “1 2. Now, ﬁnd Ex(x) using the one-dimensional divergence equation Pv - € 21 2 *03‘ ‘(0‘ ”/01 1- "l , \$0 Comte” n :4 1 / + Er: KHO‘B'W/n .9" 0<x<I E»- ‘ “MW“ k"; F" x> x A)». El: (our- - vow:- ~ . E :. -1on-v/nr lfOAN‘ﬁ-K 7") 0.? I__ _ +zolw/H " 3. Finally, using the one-dimensional form of the equation E - -VV, ﬁnd the potential (voltage) V(x) everywhere. Take x: O as the reference point for zero potential (ground) so that V(O) = 0. Do this by integrating from 0 toward positive and negative values of x. Again, you will need expressions for three regions (one for x < —T/2, one for -T/2 < x < +T/2, and one for x > T/2). Sketch your result on the graph provided. 9 v V (XI 0 04m 1- o Mk 1 ”1111411ij x xi S:L‘f 3 / )x 6 0 - POMBOL‘ K'- W Z~W03vaﬁ 3-) ; T "H‘ 2, W13: _\vo.)uv.Q.s-1o M)" k, L :—.SO3V luau» >031. 9.41m tum-77;) +u¢[email protected]»v(u)4; .____ 9 [‘;.va cw ”QC?“ wtik 5b” ~10|V~Vl ‘9' Li‘ukq + i kv/01S‘K as ‘ ‘ ésXHS‘H""') ...
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